Put in the simplest way, to calculate the power needed for a feeder drive you need to know the area of the feeder inlet, the weight of material acting through the interface and the strength of the material being handing. The last two are involved, so you had best contact a specialist like TUNRA or Jenike and Johanson. ■

My reply above may have not been too helpful so, for more background, see my comments on 'Ask Lyn' on how to assess loads on feeders. ■

Sorry, 'load on feeders' is in the 'Solid Sense' column. ■

Very good share! Very good share! Very good share! ■

Hello,

It seems, possibly you are referring to travelling feed hopper and its power calculation for travel drive.

I understand it is a travelling hopper mounted on four wheels (flanged). This is moving on rails (like railway track) and you are asking for power calculation for travel drive. Please clarify to enable me to respond your query.

Ishwar G. Mulani

Author of Book: ‘Engineering Science And Application Design For Belt Conveyors’. Conveyor design basis is ISO (thereby book is helpful to design conveyors as per national standards of most of the countries across world). New print Nov., 2012.

Author of Book: ‘Belt Feeder Design And Hopper Bin Silo’

Advisor / Consultant for Bulk Material Handling System & Issues.

Pune, India. Tel.: 0091 (0)20 25871916

Email: conveyor.ishwar.mulani@gmail.com

Website: www.conveyor.ishwarmulani.com ■

Hello,

Referring to my earlier posting, attached sketch shows the likely arrangement of pertaining bulk material handling system.

Ishwar G. Mulani

Author of Book: ‘Engineering Science And Application Design For Belt Conveyors’. Conveyor design basis is ISO (thereby book is helpful to design conveyors as per national standards of most of the countries across world). New print Nov., 2012.

Author of Book: ‘Belt Feeder Design And Hopper Bin Silo’

Advisor / Consultant for Bulk Material Handling System & Issues.

Pune, India. Tel.: 0091 (0)20 25871916

Email:

href="mailto:conveyor.ishwar.mulani@gmail.com">conveyor.ishwar.mulani@gmail.com

Website:

href="http://www.conveyor.ishwarmulani.com" target="blank">www.conveyor.ishwarmulani.com

Attachments

shiftable hopper.jpg (JPEG)

■

Hello,

As an example full wagon rake or segment of wagon rake motional resistance is comprising of following resistances:

1) Wagon rolling resistance (wheeled wagon friction resistance)

2) Gravity resistance due to up-gradient or down-gradient of rail track (slope resistance)

3) Wind resistance

For explanation purpose, following technical features are considered here:

- Wagon has 4 wheels or 8 wheels construction (i.e. whether usual size or double size)

The construction considers 150 mm diameter shaft for the wheels, 140 mm diameter bearing bore, and wagon wheel diameter 750 mm. The wheel has firm grip on shaft ends of 140 mm diameter to make it single assembly (that is 150 mm diameter shaft is the shaft portion between two bearings of 140 mm diameter). The shaft along with the inner race rotates in bearings.

1) Friction resistance: As per SKF bearings catalogue; for the aforesaid type arrangement and double row spherical roller bearing, the friction coefficient with respect to its own bore is 0.0018.

Hence for the wagon: Rolling friction coefficient is 0.0018 x (140 Divided By 750) = 0.000336

So as an example, if we consider loaded one wagon weight as 90 tonne, then its rolling friction resistance is 90000 kgf x 0.000336 = 30.24 kgf (that is 296.7 N) per wagon.

2) Gravity resistance: The rail track will not be 100% horizontal all the time (due to constructional tolerance or land settlement). I think 100 mm uprise or decline in 10 m horizontal distance would be a fair judgment. So the gravity resistance for pulling one wagon will be as below:

Wagon mass M x (Gradient that is 100 Divided By 10000) = 90000 x 0.01 = 900 kgf (because the angle is very small, sin and tan will be practically same). Its value in N is 8829.

3) Wind resistance: As for the wind resistance, one can refer to the structural design engineer who regularly calculates wind thrust on buildings / steel structures and so on. The exercise might involve wind effect in varied directions of the wind flow that is wind striking from front end or back end or on sides at 45 degree angle to rake length, etc.

Ishwar G. Mulani

Author of Book: ‘Engineering Science And Application Design For Belt Conveyors’. Conveyor design basis is ISO (thereby book is helpful to design conveyors as per national standards of most of the countries across world).

Author of Book: ‘Belt Feeder Design And Hopper Bin Silo’

Advisor / Consultant for Bulk Material Handling System & Issues.

Pune, India. Tel.: 0091 (0)20 25871916

Email: conveyor.ishwar.mulani@gmail.com

Website: www.conveyor.ishwarmulani.com ■

Hello,

In the earlier posting, I have given solution considering spherical roller bearings supporting the rotating shaft of the wagon.

Here is the other possibility, wherein there is bronze bush bearing (instead of spherical roller bearings) supporting each end of the rotating shaft. The bush bearings are to be appropriately lubricated all the time. Dry situation is not occurring . The wagon development might have passed from such situation.

The friction coefficient between bronze bush and steel shaft (lubricated) is mu = 0.20. The shaft diameter in bearing is 140 mm as before and wheel diameter is 630 mm as before.

The effective rolling friction coefficient is 0.20 Multiplied by (140 Divided By 630) = 0.0374 say 0.038. Considering weight features same as before, the wagon rolling friction resistance is 0.038 Multiplied By 90000 = 3420 kgf per wagon (as against 30.24 kgf with antifriction spherical roller bearings).

The other resistances such as climb (up or down), wind, etc. will remain as before.

Ishwar G. Mulani

Author of book: Engineering Science And Application Design For Belt Conveyors

Author of book: Belt Feeder Design And Hopper Bin Silo

Advisor / Consultant for Bulk Material Handling System & Issues.

Pune, India.

Tel. no.: 0091 (0)20 2587 1916

Email: conveyor.ishwar.mulani@gmail.com

Website: www.conveyor.ishwarmulani.com (Also to see available high class design programs)

Interestingly, one more posting on the subject follows, in a week. ■

Hello,

The calculation for wagon rolling resistance is the calculation at mega scale.

The example of such calculation at small scale happens to be for calculation of conveying friction coefficient for apron pan or chain conveyor (akin to belt conveyor conveying friction coefficient say 0.025 to 0.03, etc.).

In the following calculation for apron pan conveyor (chain conveyor, etc.) it is 0.2. Chain has two side links which are kept in position by force fitted pin of say 25 mm diameter. On this pin there is a roller say of 25 mm inner diameter and 50 mm outside diameter which is free to rotate as per conveyor speed. For steel pin and steel roller the friction coefficient with respect to steel versus steel (not lubricated) is 0.40. So rolling friction coefficient is 0.4 Multiplied By 25 Divided By 50 = 0.2 (as against belt conveyor conveying friction coefficient of 0.02 or 0.025).

So our forefathers in ancient times kept the cart / chariot wheel diameter quite big compared to small diameter of axle for easy / speedy pulling of the cart by bullocks / horses.

Ishwar G. Mulani

Author of book: Engineering Science And Application Design For Belt Conveyors

Author of book: Belt Feeder Design And Hopper Bin Silo

Advisor / Consultant for Bulk Material Handling System & Issues.

Pune, India.

Tel. no.: 0091 (0)20 2587 1916

Email: conveyor.ishwar.mulani@gmail.com

Website: www.conveyor.ishwarmulani.com (Also to see available high class design programs) ■